Animated Solution for Physics - Current Electricity: On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is $1\text{k}\Omega$. How much was the resistance on the left slot before interchanging the resistances?

The meter bridge is a classic application of the Wheatstone bridge principle, allowing us to measure unknown resistances with remarkable precision. In this problem, we are presented with a fascinating scenario where interchanging the resistances causes a measurable shift in the balance point. Let's dive into the mechanics of this shift and unravel the math behind it.

Imagine a standard meter bridge. We have two unknown resistances, $X$ and $Y$, connected in the left and right gaps, respectively. The problem provides a crucial piece of information: the resistance of their series combination is $1\text{k}\Omega$.

This immediately gives us our first constraint equation:

This means that if we can find one resistance, the other is simply $1000$ minus that value.

The meter bridge operates on the principle of a balanced Wheatstone bridge. When the galvanometer shows zero deflection, the ratio of the resistances in the gaps is exactly equal to the ratio of the corresponding balancing lengths on the meter wire.

For our initial setup, let the balancing length from the left end be $l$ centimeters. The remaining length on the right is naturally $100-l$. Applying the balancing condition, we get:

Substituting $Y=1000-X$, our master equation for the initial state becomes:

Now comes the twist. The resistances $X$ and $Y$ are interchanged. $X$ moves to the right gap, and $Y$ moves to the left gap. The problem states that this causes the balance point to shift to the left by $10\text{ cm}$.

Therefore, our new balancing length is $l-10$. Applying the balancing condition to this new configuration, we have:

Substituting $Y=1000-X$ and simplifying the denominator on the right side, we get:

Taking the reciprocal of both sides to match the structure of our first equation:

We now have two distinct expressions for the ratio $\frac{X}{1000-X}$. By equating them, we can eliminate the resistances entirely and solve for the initial balancing length $l$:

Cross-multiplying yields:

Expanding both sides:

Notice how the $l^2$ terms beautifully cancel out from both sides, leaving us with a simple linear equation:

With the initial balancing length found, we can substitute it back into our first equation to find $X$:

Cross-multiplying one last time:

$$X=550\Omega$$

The resistance in the left slot before interchanging was $550\Omega$. This problem elegantly demonstrates how physical manipulations in a circuit translate into symmetric mathematical equations, allowing us to deduce unknown values with pure logic.